Understanding the Concept:
The drain current (\(I_D\)) of an N-channel enhancement mode MOSFET operating in the saturation region (where the channel is pinched off at the drain side) is governed by the square-law characteristic equation:
\[
I_D = \frac{1}{2} \mu_n C_{ox} \frac{W}{L} (V_{GS} - V_{th})^2
\]
Where:
• \(\mu_n\) is the electron mobility in the channel.
• \(C_{ox}\) is the gate oxide capacitance per unit area (\(C_{ox} = \frac{\epsilon_{ox}}{t_{ox}}\)).
• \(W\) is the channel width.
• \(L\) is the channel length.
• \(V_{GS}\) is the applied gate-to-source voltage.
• \(V_{th}\) is the threshold voltage.
From this physical model, we observe the proportional scaling relationships between geometric dimensions and the output current capability.
Step 1: Establish proportional relations between \(I_D\) and dimensions.
Assuming that electrical bias voltages (\(V_{GS}\)) and material properties (\(\mu_n, C_{ox}\)) remain completely fixed, we can group them together as a constant \(K\):
\[
I_D = K \cdot \frac{W}{L}
\]
This reveals that:
• Drain current is directly proportional to the channel width: \(I_D \propto W\)
• Drain current is inversely proportional to the channel length: \(I_D \propto \frac{1}{L}\)
Step 2: Analyze the condition to double the current (\(I_{D2} = 2 I_{D1}\)).
Let the new length parameter be \(L'\). Setting up the ratio for changing only the channel length parameter \(L\):
\[
\frac{I_{D2}}{I_{D1}} = \frac{L}{L'}
\]
We want the current to double, meaning \(\frac{I_{D2}}{I_{D1}} = 2\):
\[
2 = \frac{L}{L'} \implies L' = \frac{L}{2}
\]
Hence, to successfully achieve double the drain current while keeping all other factors constant, the channel length must be reduced to exactly half its initial value.
Step 3: Evaluate option validity.
Let's check why the alternative statements are incorrect:
• Doubling channel length (\(L' = 2L\)) scales current by \(\frac{1}{2}\) (halves it).
• Halving channel width (\(W' = \frac{W}{2}\)) scales current by \(\frac{1}{2}\) (halves it).
• Doubling gate oxide thickness (\(t_{ox}\)) halves \(C_{ox}\), which cuts current in half.
Thus, option (C) is the only geometrically accurate solution.